Concentrations, Equilibrium Constant, and pH of Acid Base Reactions
Tedd Kourkounakis Tuesday, May 12, 2015

Up to this point, Acid Base equilibrium problems are solved the same way any other equilibrium problem we have encountered would be.

Eg) A 0.25M solution of acetic acid reacts with with water to form hydronium and acetate ions. If the Ka (Equilibrium constant of an acid) = 1.8x10-5, calculate the equilibrium concentrations of all species.

HC2H3O2 (aq) + H2O(l) à C2H3O2-(aq) + H3O+(aq)

MR	1	. - .	. 1 .	. 1 .
I	0.25	-	0	0
C	-x	-	x	x
E	0.25 - x	-	x	x

Check for approximation rule:

0.25 / 1.8x10-5 > 1000 Therefore, the approximation rule may be applied.

Ka = ([C2H3O2-][ H3O+]) / [HC2H3O2]

1.8x10-5 = ([x][x]) / [0.25 – x] Now add the approx. rule.

1.8x10-5 = [x2] / [0.25]

x = 2.1x10-3

Therefore, [HC2H3O2] = 0.25M
[ H3O+] = 2.1x10-3 M
[C2H3O2-] = 2.1x10-3 M

To solve for pH, we use the equation pH = -log[H3O+]

So, pH = -log[2.1x10-3]
pH = 2.68

If this question was instead a base, we would use the equation pOH = -log[OH-], then pH + pOH = 14 to solve for pH.

When going from concentrations to pH, the number of digits in the concentration = the number of decimal places in the pH.**

Homework

Read and take notes on pg. 492-502
Answer questions on pg. 502 #21-30 and pg. 503 #4,10

Resources
Sample Acid Base Equilibrium Step by Step Walkthrough
http://www.chemteam.info/AcidBase/Ka-Solving1.html

Strong vs. Weak Acid / Base, Conjugate Acid / Base, Ionization of Water, and Percent Ionization
Tedd Kourkounakis Wednesday, May 13, 2015

Strong Acid: Ionizes completely in water to form hydronium ions and the acid’s anion. A strong acid can be determined by using the equilibrium constant chart and looking for a Ka value of ‘very large’. (E.g. hydrochloric, nitric, sulfuric)
Weak Acid: Ionizes <50% in water to form hydronium ions and the acid’s anion (E.g. acetic, formic, nitrous)
Strong Base: A metal hydroxide that dissociates completely in water to form hydroxide and metal ions (E.g. NaOH, Ca(OH)2).
Weak Base: A substance that reacts with water to form hydroxide and cations

2 H20(l) à H30+(aq) + OH-(aq)
The equilibrium constant of water is Kw and must always equal 1.0x10-14 (at 25°C).
@25°C, Kw = [OH-][H30+]
= 1.0x10-14
This is not only true for water, but for any solution.

A conjugate acid /base pair is 2 compounds that differ by 1 proton (H+)
E.g. HC2H3O2 (aq) + H2O(l) à C2H3O2-(aq) + H3O+(aq)
The pairs would be HC2H3O2 (aq) [ACID] / C2H3O2-(aq) [CONJ. BASE]
and H2O(l) [BASE] / H3O+(aq) [CONJ. ACID]

Percent Ionization indicates the degree to which a weak acid or base ionizes in water using the following formula:
% Ionization = ([H3O+](aq) / [HA]initial) x 100%

Homework:
pg.514 questions #51-59

Resources:
Conjugate Acid Base Pairs

https://www.youtube.com/watch?v=7qBRIWSA3Yc
pH, pOH, pKa, pKb, and pKw
http://www.800mainstreet.com/acid_base/definitions-ph.html

ACID BASE TITRATION VIRTUAL LAB
Friday May 15th, 2015
By: Sara Hoang

Today’s Class:
-Used laptop to work on virtual Acid Base Titration lab
-Link to the sheet for the lab:

File Not Found

-Use this site instead of the given one in the sheet: www.chemcollective.org/vlab/vlab.php
-Note: If it does not work on Google Chrome, use internet explorer for this lab
-Note: Do not have to upgrade java, make sure to click ‘run’

Homework:
-Was to finish up to and including number six for the lab

Resources:

JFR Chemistry - Titration Review LINK: https://www.youtube.com/watch?v=aeA1sVm44FU

BUFFERS

Isabel Bhuiyan

Friday May 22, 2015.

buffers are mixtures of conjugate acid-base pairs
buffers allow solutions to resist changes in pH
they have common ions in them that act as reservoirs and help maintain a relatively constant pH
examples of buffers:
- acetic acid and sodium acetate
- formic acid and sodium formate
- ammonia and ammonium chloride
- methylamine and methyl ammonium nitrate

EXAMPLE PROBLEM

A buffer is made by mixing 0.60 mol of acetic acid and 0.50 mol of sodium acetate in 2.0 L of water.

a) Calculate the pH of the original buffer.

b) Calculate the pH after adding 0.10 mol of OH-, ignoring any volume changes.

same as decreasing [H3O+]
shift to the right
[HC2H3O2] will decrease by the same amount of OH- added (0.10 mol/2.0L = 0.05M)
therefore, [HC2H3O2] = 0.30M - 0.05M = 0.25M (new initial concentration)
[C2H3O2-] will increase by 0.05M, therefore [C2H3O2-] = 0.25 M + 0.05M = 0.30 M (new initial concentration)

c) Calculate the pH after adding 0.10 mol of H3O+, ignoring any volume changes.

shift to the left
[HC2H3O2] will increase by the same amount of H3O+ added (0.10 mol/2.0L = 0.05M)
therefore, [HC2H3O2] = 0.30M + 0.05M = 0.35M (new initial concentration)
[C2H3O2-] will decrease by 0.05M, therefore [C2H3O2-] = 0.25 M - 0.05M = 0.20 M (new initial concentration)

HOMEWORK

complete buffer problems on Acid-Base Equilibrium Problems sheet
study for the quest on Tuesday May 26
continue working on your Book ISU

RESOURCES

That Chem Guy - Buffers Acids and Bases Chemistry Tutorial https://www.youtube.com/watch?v=avnjBlicpFk
Khan Academy - Buffers https://www.youtube.com/watch?v=HzmI7A578ss

May 26th, 2015
Sara Hoang
CLASS WORK:

Did Acid-Base Equilibrium quest

HOMEWORK:

Continue working on the Book ISU (written portion due June 2nd, 2015)
If done the written portion of Book ISU then work on the presentation portion!
Review notes for Solubility Equilibrium and complete the Solubility Equilibrium worksheet

RESOURCES:

ChemistNATE - What is Ksp? (Solubility Product Constant) LINK: https://www.youtube.com/watch?v=FsU8Ss6jiKs

Unit 4C Acid Base Equilibrium